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Structure of Atom MCQs

Class 11 Chemistry — questions with answers and worked explanations.

20 free questions Class 11 Chemistry Answers + explanations No sign-up

These questions are drawn from the Pariksha Sutra question bank for Structure of Atom, part of the Class 11 Chemistry syllabus. Each one shows the correct answer and, where a method helps, the working behind it.

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Questions

Question 1
Which model of the atom is popularly called the "plum pudding" model?
  1. Quantum mechanical model
  2. Bohr's model
  3. Thomson's model
  4. Rutherford's model
Answer: Thomson's model
  1. J.J. Thomson envisioned the atom as a uniformly positive sphere in which the negatively‑charged electrons are scattered like raisins.
  2. Because the picture resembles a plum‑filled pudding, this description is popularly called the “plum pudding” model.
  3. Thus the model referred to is Thomson’s model.
Question 2
Which subatomic particle is electrically neutral?
  1. alpha particle
  2. neutron
  3. proton
  4. electron
Answer: neutron
  1. Neutron has no net electric charge, its constituent quarks’ charges cancel out.
  2. Proton contains two up quarks (+2/3 each) and one down quark (‑1/3), giving +1 charge.
  3. Electron carries a single negative charge (‑1).
  4. Alpha particle is a helium nucleus (2 protons + 2 neutrons) with a +2 charge, so the neutral particle is the neutron.
Question 3
The frequency of radiation of wavelength 600 nm is (c = 3*10^8 m/s)
  1. 2*10^14 Hz
  2. 1.8*10^14 Hz
  3. 5*10^15 Hz
  4. 5*10^14 Hz
Answer: 5*10^14 Hz
  1. Unit Conversion: Convert the wavelength from nanometers (nm) to meters (m):
  2. λ = 600 nm = 600 × 10−9 m = 6 × 10−7
  3. m
  4. Substitution: Substitute the known values into the rearranged formula:
  5. ν = (3 × 108
  6. m/s) / (6 × 10−7 m)
  7. Simplification:
  8. ν = (3 / 6) × 108 − (−7)
  9. ν = 0.5 × 1015 Hz = 5 × 1014 Hz
  10. FINAL ANSWER
  11. ν = 5.0 × 1014 Hz (or 500 THz)
Question 4
The spectral series of hydrogen lying in the visible region is the
  1. Balmer series
  2. Paschen series
  3. Pfund series
  4. Lyman series
Answer: Balmer series
  1. Visible light corresponds to wavelengths roughly 400–700 nm.
  2. In hydrogen, transitions that emit photons in this range end at the n=2 level.
  3. These transitions are described by the Balmer formula (R_H(1/2²–1/n²)).
  4. Hence the visible spectral series of hydrogen is the Balmer series.
Question 5
The radius of the second Bohr orbit (n=2) of hydrogen is
  1. 2.116 Angstrom
  2. 4.761 Angstrom
  3. 1.058 Angstrom
  4. 0.529 Angstrom
Answer: 2.116 Angstrom
  1. Bohr radius a₀ = 0.529 Å for n = 1.
  2. In Bohr model, radius rₙ = n²·a₀.
  3. For the second orbit n = 2, r₂ = 2² × 0.529 Å = 4 × 0.529 Å.
  4. 2.116 Angstrom
Question 6
The uncertainty principle is significant only for
  1. charged bodies
  2. planets
  3. microscopic particles
  4. macroscopic objects
Answer: microscopic particles
  1. Δx·Δp ≥ ħ/2 shows position–momentum uncertainty grows as the particle’s wavelength becomes comparable to its size.
  2. For macroscopic bodies the de Broglie wavelength is extremely tiny, making Δx and Δp practically zero.
  3. Thus the principle is noticeable only for microscopic particles. microscopic particles
Question 7
The relative electric charge of a proton is
  1. +1
  2. +2
  3. 0
  4. -1
Answer: +1
  1. Relative charge is a dimensionless number that expresses the electric charge of a subatomic
  2. particle normalized against the magnitude of the elementary charge (e), where:
  3. e ≈ 1.602 × 10−19 C
  4. Since the proton carries exactly one elementary positive charge, its relative charge is evaluated as:
  5. Relative Charge = +e / e = +1
Question 8
The wavenumber of radiation is defined as
  1. lambda/c
  2. 1/lambda
  3. c/lambda
  4. nu*lambda
Answer: 1/lambda
  1. Wavenumber is defined as the number of wave cycles in a unit length.
  2. It is the reciprocal of the wavelength because wavelength is the distance for one cycle.
  3. Thus wavenumber = 1 divided by λ, i.e., 1/λ.
Question 9
The minimum energy required to eject an electron from a metal surface is called the
  1. work function
  2. kinetic energy
  3. binding energy
  4. ionization energy
Answer: work function
  1. In the photoelectric effect a photon must supply enough energy to free an electron from the metal.
  2. That threshold energy is defined as the work function φ, the smallest energy needed to remove an electron.
  3. Therefore the minimum energy required to eject an
Question 10
The radius of the first Bohr orbit of hydrogen is approximately
  1. 2.12 Angstrom
  2. 0.529 Angstrom
  3. 1.06 Angstrom
  4. 4.76 Angstrom
Answer: 0.529 Angstrom
  1. Bohr radius a₀ = ħ²/(mₑ e²) = 0.529 × 10⁻¹⁰ m.
  2. Convert to Å (1 Å = 10⁻¹⁰ m) gives a₀ = 0.529 Å.
  3. Thus the radius of the first Bohr orbit of hydrogen is 0.529 Angstrom.
Question 11
The mathematical form of the Heisenberg uncertainty principle is
  1. delta x * delta p <= h/4pi
  2. delta x * delta p = h/2pi
  3. delta x * delta p >= h/4pi
  4. delta x * delta p = h
Answer: delta x * delta p >= h/4pi
  1. Uncertainty principle relates the spread in position (Δx) and momentum (Δp) of a particle.
  2. Using Planck’s constant h, the reduced constant is ħ = h/2π, and the principle states Δx·Δp ≥ ħ/2.
  3. Substituting ħ gives Δx·Δp ≥ (h/2π)/2 = h/4π.
  4. Thus the correct mathematical form is delta x * delta p >= h/4pi
Question 12
The subshell designation for l=2 is
  1. s
  2. d
  3. f
  4. p
Answer: d
  1. The azimuthal quantum number l determines the subshell type: l = 0 → s, l = 1 → p, l = 2 → d, l = 3 → f.
  2. For the given value l = 2, the corresponding subshell is the d‑type.
  3. Hence the subshell designation is d.
Question 13
In Rutherford's alpha-particle scattering experiment, most alpha particles passed straight through the foil undeflected because
  1. the atom is largely empty space
  2. the nucleus repelled all of them
  3. the alpha particles were neutral
  4. electrons attracted them strongly
Answer: the atom is largely empty space
  1. Alpha particles are positively charged and experience strong repulsion only when they come very close to the dense, positively charged nucleus.
  2. The nucleus occupies an extremely small volume compared with the overall size of the atom.
  3. Therefore, most alpha particles travel through regions where there is essentially no material to deflect them.
  4. Hence most alpha particles passed straight through the foil because the atom is largely empty space.
Question 14
From his scattering experiment, Rutherford concluded that the nucleus is
  1. negatively charged
  2. large and light
  3. spread throughout the atom
  4. small and positively charged
Answer: small and positively charged
  1. Alpha particles mostly passed through the foil undeviated, showing most of the atom is empty space.
  2. A few particles were sharply deflected, indicating a concentrated region of large positive charge.
  3. The deflection angles required this region to be very small compared with the atomic size.
  4. Thus Rutherford concluded the nucleus is small and positively charged.
Question 15
The wavenumber of radiation of wavelength 500 nm is
  1. 5*10^6 m^-1
  2. 5*10^7 m^-1
  3. 2*10^6 m^-1
  4. 2*10^7 m^-1
Answer: 2*10^6 m^-1
  1. Wavenumber (ṽ) is defined as the reciprocal of wavelength: ṽ = 1/λ.
  2. Convert 500 nm to metres: 500 nm = 500 × 10⁻⁹ m = 5 × 10⁻⁷ m.
  3. Take the reciprocal: 1/(5 × 10⁻⁷ m) = (1/5) × 10⁷ m⁻¹ = 0.2 × 10⁷ m
Question 16
The Lyman series of the hydrogen spectrum lies in the
  1. ultraviolet region
  2. visible region
  3. infrared region
  4. microwave region
Answer: ultraviolet region
  1. Transition to the ground level n=1 releases the largest energy among hydrogen series.
  2. Using ΔE = hc/λ, the high ΔE gives λ < 400 nm.
  3. Thus the Lyman series lies in the ultraviolet region.
Question 17
As the value of n increases, the energy difference between successive orbits
  1. decreases
  2. remains constant
  3. becomes zero
  4. increases
Answer: decreases
  1. Energy of the nth orbit is En = –13.6 eV / n² (hydrogen‑like atom).
  2. The gap to the next level is ΔE = En+1 – En = 13.6 (1/n² – 1/(n+1)²).
  3. Since 1/n² – 1/(n+1)² gets smaller for larger n, the spacing shrinks.
  4. Thus the energy difference between successive orbits decreases.
Question 18
The uncertainty principle rules out the existence of
  1. wavefunctions
  2. definite fixed orbits
  3. quanta
  4. orbitals
Answer: definite fixed orbits
  1. Heisenberg’s principle states Δx·Δp ≥ ħ/2, so an electron’s position and momentum cannot be known simultaneously with arbitrary precision.
  2. If an electron were in a definite fixed orbit, both its radius (position) and speed (momentum) would be exact, violating the principle.
  3. Therefore the uncertainty principle rules out the existence of definite fixed orbits.
Question 19
According to Bohr, the angular momentum of an electron in an orbit is an integral multiple of
  1. h/2pi
  2. h*nu
  3. pi*h
  4. h/pi
Answer: h/2pi
  1. Bohr postulated that the electron’s orbital angular momentum is quantized.
  2. He expressed this quantization as L = n·h/2π, where n is an integer (principal quantum number).
  3. Thus each allowed orbit has angular momentum equal to an integral multiple of h/2π.
  4. Hence the correct choice is h/2π
Question 20
The number of protons in the nucleus of an atom is called its
  1. isotope number
  2. atomic number
  3. neutron number
  4. mass number
Answer: atomic number
  1. Protons determine the identity of an element.
  2. The count of protons in the nucleus is defined as the atomic number.
  3. Atomic number distinguishes one element from another.
  4. Thus the number of protons in the nucleus of an atom is called its atomic number.

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